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A 46kg sample of water absorbs 4.00×102kJ of heat. If the water was initially at 29.7 ∘ C , what is its final temperature?
DrBob222
answered
11 years ago
11 years ago
Explain Bot
answered
11 months ago
11 months ago
To find the final temperature of the water, we need to use the specific heat formula:
q = mcΔT
Where:
q is the amount of heat absorbed (4.00×10^2 kJ)
m is the mass of the water (46 kg)
c is the specific heat capacity of water
ΔT is the change in temperature
The specific heat capacity of water is approximately 4.18 kJ/(kg⋅°C).
Rearranging the formula, we have:
ΔT = q / (mc)
Substituting the given values, we have:
ΔT = (4.00×10^2 kJ) / (46 kg × 4.18 kJ/(kg⋅°C))
Now let's calculate:
ΔT = (4.00×10^2) / (46 × 4.18)
ΔT ≈ 2.941 °C
To find the final temperature, we add the change in temperature to the initial temperature:
Final temperature = Initial temperature + ΔT
Final temperature = 29.7 °C + 2.941 °C
Final temperature ≈ 32.641 °C
Therefore, the final temperature of the water is approximately 32.641 °C.